## Tag `00E8`

Chapter 10: Commutative Algebra > Section 10.16: The spectrum of a ring

Lemma 10.16.10. Let $R$ be a ring. The space $\mathop{\rm Spec}(R)$ is quasi-compact.

Proof.It suffices to prove that any covering of $\mathop{\rm Spec}(R)$ by standard opens can be refined by a finite covering. Thus suppose that $\mathop{\rm Spec}(R) = \cup D(f_i)$ for a set of elements $\{f_i\}_{i\in I}$ of $R$. This means that $\cap V(f_i) = \emptyset$. According to Lemma 10.16.2 this means that $V(\{f_i \}) = \emptyset$. According to the same lemma this means that the ideal generated by the $f_i$ is the unit ideal of $R$. This means that we can write $1$ as afinitesum: $1 = \sum_{i \in J} r_i f_i$ with $J \subset I$ finite. And then it follows that $\mathop{\rm Spec}(R) = \cup_{i \in J} D(f_i)$. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 3140–3143 (see updates for more information).

```
\begin{lemma}
\label{lemma-quasi-compact}
Let $R$ be a ring. The space $\Spec(R)$ is quasi-compact.
\end{lemma}
\begin{proof}
It suffices to prove that any covering of $\Spec(R)$
by standard opens can be refined by a finite covering.
Thus suppose that $\Spec(R) = \cup D(f_i)$
for a set of elements $\{f_i\}_{i\in I}$ of $R$. This means that
$\cap V(f_i) = \emptyset$. According to Lemma
\ref{lemma-Zariski-topology} this means that
$V(\{f_i \}) = \emptyset$. According to the
same lemma this means that the ideal generated
by the $f_i$ is the unit ideal of $R$. This means
that we can write $1$ as a {\it finite} sum:
$1 = \sum_{i \in J} r_i f_i$ with $J \subset I$ finite.
And then it follows that $\Spec(R)
= \cup_{i \in J} D(f_i)$.
\end{proof}
```

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